(i) We argue by induction on \(c=\mbox{cx}\,_R(M)\) . If \(c=0\)  then \(\mbox{Pd}\,_R(M)<\infty \)  and the formula holds by Auslander's original result, so suppose that \(c>0\)  and that \(\eta \in \mbox{Ext}\,^*_R(M,M)\)  reduces the complexity of \(M\) .
The exact sequence \(0\rightarrow M\rightarrow K_{\eta }\rightarrow \Omega ^n_R(M)\rightarrow 0\) , induces a long exact sequence
\(\cdots \rightarrow \mbox{Tor}\,_i^R(M,N)\rightarrow \mbox{Tor}\,_i^R(K_{\eta },N)\rightarrow \mbox{Tor}\,_{i+n}^R(M,N)\rightarrow \cdots \) 
of homology modules. Therefore \(\mbox{Tor}\,_i^R(K_{\eta },N)=0\)  for all \(i>0\) . As \(\mbox{cx}\,_R(K_{\eta })<c\)  and \(\mbox{G-dim}\,_R(K_{\eta })<\infty \) ,
\(\mbox{depth}\,_R(K_{\eta })+\mbox{depth}\,_R(N)=\mbox{depth}\,_R(K_{\eta }\otimes _RN)+\mbox{depth}\,R\) 
by induction hypothesis.
Now by induction on \(\mbox{G-dim}\,_R(M)\) , we show that the formula holds. If \(\mbox{G-dim}\,_R(M)=0\)  then by the Lemma REF , \(\mbox{Ext}\,^i_R(\mbox{Tr}\,M,N)=0\)  for all \(i>0\) .
Hence by Theorem REF , \(M\otimes _RN\cong \mbox{Hom}\,_R(M^*,N)\)  and also by the exact sequence (REF ), \(\mbox{Ext}\,^i_R(M^*,N)=0\)  for all \(i>0\) .
Therefore by [1]}, \(\mbox{depth}\,_R(M\otimes _RN)=\mbox{depth}\,_R(\mbox{Hom}\,_R(M^*,N))=\mbox{depth}\,_R(N)\)  and so by the Auslander-Bridger formula,
\(\mbox{depth}\,_R(M)+\mbox{depth}\,_R(N)=\mbox{depth}\,_R(M\otimes N)+\mbox{depth}\,R\) .
Now let \(\mbox{G-dim}\,_R(M)>0\) , if \(n=0\)  then we obtain the following exact sequence
\(0\rightarrow M\otimes _RN\rightarrow K_{\eta }\otimes _RN\rightarrow M\otimes _RN\rightarrow 0.\) 
As \(\mbox{G-dim}\,_R(M)>0\) , \(\mbox{G-dim}\,_R(\Omega _R(M))=\mbox{G-dim}\,_R(M)-1\) . Note that by the proof of [2]}, \(\Omega _R(M)\)  has also reducible complexity. Therefore,
\(\mbox{depth}\,_R(\Omega _R(M))+\mbox{depth}\,_R(N)=\mbox{depth}\,_R(\Omega _R(M)\otimes N)+\mbox{depth}\,R\) 
by induction hypothesis.
From the exact sequence \(0\rightarrow \Omega _R(M)\rightarrow F\rightarrow M\rightarrow 0\) , where \(F\)  is a free module, we obtain the exact sequence
\(0\rightarrow \Omega _R(M)\otimes _RN\rightarrow F\otimes _RN\rightarrow M\otimes _RN\rightarrow 0\) . Therefore, \(\mbox{depth}\,_R(M\otimes _RN)\ge \min \lbrace \mbox{depth}\,_R(N),\mbox{depth}\,_R(\Omega _R(M)\otimes _R N)-1\rbrace \) , by the depth Lemma. Now by (REF ), \(\mbox{depth}\,_R(\Omega _R( M)\otimes _RN)-1=\mbox{depth}\,_R(N)-\mbox{G-dim}\,_R(M)<\mbox{depth}\,_R(N)\)  and so \(\mbox{depth}\,_R(M\otimes _RN)\ge \mbox{depth}\,_R(N)-\mbox{G-dim}\,_R(M)\) . On the other hand, if \(\mbox{depth}\,_R(M\otimes _RN)>\mbox{depth}\,_R(N)-\mbox{G-dim}\,_R(M)\)  then by the exact sequence (REF ), it is obvious that \(\mbox{depth}\,_R(K_{\eta }\otimes _RN)>\mbox{depth}\,_R(N)-\mbox{G-dim}\,_R(M)=\mbox{depth}\,_R(N)-\mbox{G-dim}\,_R(K_{\eta })\) , which is a contradiction by (REF ). Hence
\(\mbox{depth}\,_R(M)+\mbox{depth}\,_R(N)=\mbox{depth}\,_R(M\otimes N)+\mbox{depth}\,R\) .
Now let \(n>0\) , then \(\mbox{G-dim}\,_R(\Omega ^n_R(M))=\max \lbrace 0,\mbox{G-dim}\,_R(M)-n\rbrace <\mbox{G-dim}\,_R(M)\) . Note that by the proof of the [2]}, \(\Omega ^n_R(M)\)  has also reducible complexity and so by induction hypothesis, \(\mbox{depth}\,_R(\Omega ^n_R(M)\otimes _RN)=\mbox{depth}\,_R(N)-\mbox{G-dim}\,_R(\Omega ^n_R(M))\) . Therefore, as \(\mbox{G-dim}\,_R(M)=\mbox{G-dim}\,_R(K_{\eta })\) , \(\mbox{depth}\,_R(K_{\eta }\otimes _RN)<\mbox{depth}\,_R(\Omega ^n_R(M)\otimes _RN)\)  by (REF ) and so from the exact sequence \(0\rightarrow M\otimes _RN\rightarrow K_{\eta }\otimes _RN\rightarrow \Omega ^n_R(M)\otimes _RN\rightarrow 0\) , it is obvious that \(\mbox{depth}\,_R(M\otimes _RN)=\mbox{depth}\,_R(K_{\eta }\otimes _RN)\) .
Therefore by (REF ), \(\mbox{depth}\,_R(M)+\mbox{depth}\,_R(N)=\mbox{depth}\,_R(M\otimes N)+\mbox{depth}\,R\) .
(ii) We argue by induction on \(\mbox{cx}\,_R(M)=c\) . If \(c=0\)  then \(\mbox{Pd}\,_R(M)<\infty \)  and the formula holds by Auslander's original result, so suppose that \(c>0\)  and that \(\eta \in \mbox{Ext}\,^*_R(M,M)\)  reduces the complexity of \(M\) . The exact sequence \(0\rightarrow M\rightarrow K_{\eta }\rightarrow \Omega ^n_R(M)\rightarrow 0\) , induces a long exact sequence
\(\cdots \rightarrow \mbox{Tor}\,_q^R(M,N)\overset{f}{\rightarrow }\mbox{Tor}\,_q^R(K_{\eta },N)\rightarrow \mbox{Tor}\,_{q+n}^R(M,N)\rightarrow \cdots \) 
of homology modules.
From (REF ), it is obvious that \(q=\sup \lbrace i\mid \mbox{Tor}\,_i^R(K_{\eta },N)\ne 0\rbrace \) . If \(n=0\)  then from (REF ), we obtain the following exact sequences
\(0\rightarrow \mbox{Tor}\,_q^R(M,N)\rightarrow \mbox{Tor}\,_q^R(K_{\eta },N)\rightarrow \mbox{Coker}\,(f)\rightarrow 0,\) 
\(0\rightarrow \mbox{Coker}\,(f)\rightarrow \mbox{Tor}\,_q^R(M,N).\) 
If \(\mbox{depth}\,_R(\mbox{Tor}\,_q^R(M,N))=0\) , then by the exact sequence (REF ), \(\mbox{depth}\,_R(\mbox{Tor}\,_q^R(K_{\eta },N))=0\) . As \(\mbox{cx}\,_R(K_{\eta })<c\)  , by induction hypothesis, \(\mbox{depth}\,_R(N)+\mbox{depth}\,_R(K_{\eta })=\mbox{depth}\,R-q\)  and since \(\mbox{depth}\,_R(K_{\eta })=\mbox{depth}\,_R(M)\) , we are done. If \(\mbox{depth}\,_R(\mbox{Tor}\,_q^R(M,N))=1\)  then by the exact sequence (REF ), \(\mbox{depth}\,_R(\mbox{Coker}\,(f))>0\)  and so by the exact sequence (REF ), \(\mbox{depth}\,_R(\mbox{Tor}\,_q^R(K_{\eta },N))=1\) . Therefore by induction hypothesis, \(\mbox{depth}\,R+1-q=\mbox{depth}\,_R(K_{\eta })+\mbox{depth}\,_R(N)=\mbox{depth}\,_R(M)+\mbox{depth}\,_R(N)\) .
Now suppose that \(n>0\) , then from the exact sequence (REF ), it is obvious that \(\mbox{Tor}\,_q^R(M,N)\cong \mbox{Tor}\,_q^R(K_{\eta },N)\)  and so by induction hypothesis, we are done.
